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Why do planes not fly "backwards" (up from the planes POV) when in a vertical climb?
As suggested by a user in this question, I'd like to ask this question:
One of the most basic aerobatic maneuvers is a vertical climb. This can be achieved by just pitching up to what appears to be 90°.
Assuming it is really a 90° pitch up maneuver, wouldn't the lift of the wing (though smaller due to less speed) and the missing gravity to counteract the lift make the plane go "up" (again, planes point of view!)?

Short answer: Yes. But only for a moment.
As you suggest, let's focus on the 90° case, because many things will become simpler. Intermediate climb angles can be covered by a linear superposition of horizontal and vertical flight.
Now we shoot up at 90° flight path angle (engine thrust permitting) and leave some angle of attack for the wing to create lift. Lift acts orthogonally to the local flow direction, so the lift will pull the aircraft sideways (up in the aircraft's frame of reference). Since there is no other force which will oppose this lift (drag, weight and thrust all act at 90° to the lift, parallel to the flight path), the aircraft will accelerate sideways (again, up in the aircraft's frame of reference). The angle of attack will immediately decrease and lift creation will stop — or the pilot forces the angle of attack to stay constant by pulling the elevator, and now the aircraft will further increase its flight path angle.
That is a complicated way of saying that the pilot forces the airplane into a loop. If he wants to maintain this 90° flight path angle, he better avoid creating any lift.

I think the short version of this is: In order to fly vertically, the pilot doesn't have to aim the thrust vector (and therefore airplane) straight up. Due to some leftover wing lift, slightly less than 90° is sufficient.

@JulianHzg: No, the short version is: Any lift will change the angle of attack until none of the lift is left, when the pilot points the airplane straight up.

Well, yes, those are equivalent. It's just the differece between the pilot wanting to fly nose-up or vertical trajectory.

Answering your question, what will happen with the list depends on what maneuver the pilot is trying to do.
In the situation you are describing and taking the picture that @PeterKämpf prepared:

In your situation the gamma angle is 90º, so basically the gravity (mg) will be aligned with drag and thrust (T) will be used to compensate both weight (mg) and drag (D). Acceleration of the airplane will be T-mg-D = ma (our friend Newton). Actually, in a typical airplane, drag will be negligible compared to weight.
In this situation, there is one variable missing, the angle of attack (alpha).
Now let's go to the maneuver...
1) I would like to have my airplane stopped in the air. In this situation you try to keep the airplane at 0 speed so you don't have any lift. I have seen this done for a few seconds (or less than one) by a Typhoon in an airshow (was about 70-80º actually). It was really difficult for the pilot to keep it stable, it was moving a little bit normal to the wing.
2) You would like to have an stable climbing totally vertical. First thing is that this is not the most efficient way to do it as it is not something normal in operation. Will be the same situation as before, but there will be some speed, the pilot will try to keep the angle of attack stable to produce no lift. Problem is that such angle of attack varies with speed... so he will struggle. The Typhoon made that to scape back from the situation... he moved up and pitched down. Also climbing intercept mission in combat airplane is close to this one still very extreme.
3) You are making a loop. You create some lift to make it possible so the airplane is turning. That's better explained by @PeterKämpf
As a general comment those maneuver are usually performed in airshow by airplanes prepared for this and I have never seen any "pure" situation of what you are describing, just situations very close to what you describe. Neither combat airplanes are using it regularly although might have the capability.
However those operations are not affordable by civil airplanes. Usually thurst compensates drag and lift weight, usually lift is of the order of 10 times bigger than drag, so basically engine thrust capability is around tenth of the weight (if you consider extreme operations you will end up with a number of 2-3 times smaller). Just take the example of A320, A318 OEW is aroudn 40Tons and the thrust is 210KN (weight is twice the maximum) and going to MTWO we have 68tons (3 times).
(remember the g on previous calculation)
So... normal planes usually don't have a big engine compared to their weight, however the Typhoon has a ratio of 1.15 thrust/weight.

The Typhoon had stability problems because it wasn't designed to do that - no low-speed attitude control. Those high-thrust maneuvers are airshow stunts only. A Harrier is designed to do that, and can park itself 50 meters up until it runs out of fuel. Harriers used to do 'Moonshot' takeoffs at airshows: vertical takeoff, rotate the plane up and the thrust down transitioning to a vertical climb. Noisy, spectacular, and a great way to empty the tanks quickly.

The Typhoon did it during a very small period. Actually I have never seen a harrier doing that, neither in an airshow (which does not mean they can simply I ddi not have the opportunity).

I once saw an F-14 do a transition from level to vertical flight popping on the afterburners just before or during the transition.  It was pretty spectacular.  The Harriers on the ship I was on always took off using the small carrier deck (LHA-4, Tarawa class) but always landed vertically.

With regards to the situation where the airplane is "suspended" in a near vertical orientation. Expanding upon what Trebia Project mentioned, as well as some of the comments I read, there are tactical situations where this sort of maneuver might be used.
We went out to the area to engage two F-16's in air-to-air combat. Our flight was a flight of four A7-E's. The A7-E had great fuel efficiency, and that combined with a maximum payload of 12,000 pounds of ordinance, made it a formidable attack jet. Also, although it was not supersonic, it came close to Mach 1 at sea level. Breaking the sound barrier at these altitudes comes with a heavy cost in fuel. Anybody chasing us in burner is going to be sucking at their fuel tanks.
The specifications that made the A7-E such a good attack jet, e.g. long legs to the enemy target, were also the specifications that meant it was not particularly suited for air-to-air combat. Our rule of thumb was that if a fighter got behind your 9/3 o'clock jettison your stores and turn with your adversary. If you could last 30 seconds in the fight, the fighter will have used a good portion of their fuel. Their situation is more severe if they are in burner with you on the deck. I mention this to provide some context to the engagement with the F-16's.
When we flew attack missions, depending on the threat, we came in low or high. In either case, as a flight of 4, we were in a box formation. The lead and their wing man abreast at the head of the flight, while the second division was abreast one another a mile in trail. Four aircraft each at the corner of a square a mile a side. This mission was to help us train to maintain flight integrity while engaging multiple fighters. From the flight I learned it takes a lot of discipline.
In the area we soon found ourselves engaged with the F-16's. We were outmatched  quickly and our box formation degraded into a circle with an A7 at each 90 degrees on the arc. Each A7 was protecting the six of the guy in front. A defensive position, without much chance to attack. I often took aircraft, with similar performance characteristics, vertically in a dog fight. The A7 can't maintain that geometry very long, and it becomes a game of chicken at some point. Who is willing to get to 0 airspeed with the nose at 90 degrees?
How did the F-16's handle the defensive position we took up? These guys both went vertical and sat above the circle using their burners to keep themselves "suspended." They were between 70 and 90 degrees, and looked like they were slowly falling forward when they would hit the burner and come back to a mostly stationary position. It was the coolest thing to watch. At some point one F-16 would drop in on our circle and try to get on the six of an A7. Of course the A7 in trail would force themselves around to an AIM-9L shot. With the increasing threat the F-16 would pull back up vertical, and the other would drop in on a different part of the circle. There was always one F-16 above the circle using their burner to stay atop the fight. They knew our energy was too low to get our nose much above the horizon.
Eventually the F-16 flight called the fight off with no kill. When we returned home and were debriefing the flight, the lead, who was a decorated Vietnam attack pilot, said "Yeah this use to happen in fights when I flew in Vietnam."

Neat story.  It's not the crate but the man sitting in it that counts.

For the plane to go straight up, there must be a force in equilibrium with the force (Lift) of the wing.  That is created with the engine (Thrust) slightly off vertical in the opposite direction.

could you please expand your reasoning? I am not sure I completely follow you.


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Can I do a dual VOR check on the ground?
I know that the regs are ambiguous on the subject, but does anyone have a good argument as to why or why not you can do a dual VOR check on the ground if the VOR signal is good?
For my particular situation, I'm at KRDM which is about six miles from the DSD vortac. There is a clear line of sight to the station, but the airport is a couple hundred feet lower than the VOR.
Seems to me that a dual ground check would be fine since no flags are visible and the station can be identified, but I'm not sure.

There are lots of airports with a VOR on the field that have designated VOR checkpoints. I don’t see why checking dual VORs with a good signal would be a problem.

Be careful performing a dual VOR check on the ground, especially if you are located near other aircraft or buildings that may interfere with the signal.  I have performed a VOR check on the ground at a few airports around the Southwest, and I have found that you can get erroneous readings, on the ground, compared to performing the dual check in the air. I would say its more accurate in the air.
As for how to log it. DATE, PLACE, ERROR, and SIGNATURE. Place, meaning the VOR station identifier. 91.171(d)

I second this. While on ground, you are subjected to all kinds of disturbances, so while it's possible, I would advice against it.

As long as the signal strength is solid, and you have a known location, with no outstanding navigational NOTAMS associated with it, there is no reason you cannot.

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What's the minimum allowed flying height of a light aircraft over a town?
I mean a Light aircraft, in Germany and a town like Heidelberg. And I mean not landing, of course, but overflying.

In the USA, the FAA rules require roughly 300 meters above the highest obstacle that is within 600 meters horizontally.  (1000 feet above an obstacle within 2000 feet)

In Europe this is regulated by SERA.5005 Visual flight rules (source, emphasis mine):

(f) Except when necessary for take-off or landing, or except by permission from the competent authority, a VFR flight shall not be flown:
(1) over the congested areas of cities, towns or settlements or over an open-air assembly of persons at a height less than 300 m (1 000 ft) above the highest obstacle within a radius of 600 m from the aircraft;
(2) elsewhere than as specified in (1), at a height less than 150 m (500 ft) above the ground or water, or 150 m (500 ft) above the highest obstacle within a radius of 150 m (500 ft) from the aircraft.

EASA has published an easy access version of the Standardised European Rules of the Air (SERA) here.
To answer your question, see: SERA.5005(f) Visual flight rules (presuming you are asking about VFR flights):

(f) Except when necessary for take-off or landing, or except by permission from the competent authority, a VFR flight shall not be flown:
(1) over the congested areas of cities, towns or settlements or over an open-air assembly of persons at a height less than 300 m (1 000 ft) above the highest obstacle within a radius of 600 m from the aircraft;

Notice also SERA.3105 Minimum heights:

Except when necessary for take-off or landing, or except by permission from the competent authority, aircraft shall not be flown over the congested areas of cities, towns or settlements or over an open- air assembly of persons, unless at such a height as will permit, in the event of an emergency arising, a landing to be made without undue hazard to persons or property on the surface. The minimum heights for VFR flights shall be those specified in SERA.5005(f) and minimum levels for IFR flights shall be those specified in SERA.5015(b).

... and GM1 SERA.3105 Minimum heights:

In cases where it is considered that the minimum heights specified in SERA.5005 and SERA.5015 are not sufficient, the competent authority may establish appropriate structures, such as controlled, restricted or prohibited airspace, and define specific conditions through national arrangements. In all cases, the related Aeronautical Information Publication (AIP) and charts should be made easy to comprehend for airspace users.

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Can I get a sport pilot license while taking Mirapex?
May I get a sport pilot license while taking the medication Mirapex for Restless Leg Syndrome?

What research have you done beforehand?

Welcome to aviation.SE! For questions about rules and regulations, please always tell us which country (or set of regulations) you're asking about. Are you asking about FAA rules?

Assuming you have never applied for a medical certificate:
You need to ask your doctor. Sport pilots don't need to have a medical certificate, but they do need to be capable of operating an airplane safely. Mirapex is on the do not issue, do not fly list, so the presumption is that you cannot fly while taking it. In order to legally fly solo or with passengers while taking Mirapex, your doctor must tell you that it's safe to do so.
Your doctor may tell you that it's not safe to fly while taking Mirapex. In that case, you only have two options: stop taking the medication (with the approval and supervision of your doctor), or don't fly (except as a student receiving instruction).
If you have applied for a medical certificate ever in your life: You need to get a medical certificate in order to fly solo or with passengers. In this case, you need to talk to an aviation medical examiner for advice; there's no way around this.

Your last paragraph is slightly incorrect: a private pilot (for example) can exercise sport privileges without a current medical. However, if their previous medical was denied or revoked then they can't fly under sport rules. AOPA has some useful information.

@Pondlife Thanks for the feedback. Looks like I was assuming that the asker has never been issued a medical certificate, and I forgot about the case where the asker has been issued a certificate but the certificate has lapsed. I should edit this answer to add more information about that.


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Is an A&P certificate ever required to work on military aircraft?
Are military aviation mechanics - including civilian contractors - ever required to obtain an A&P certificate to work on active military aircraft?

Not usually, A&P is an FAA regulation/certification, and the US Military is not required to abide by FAA regulations. Even military pilots don't (always) have pilots licenses (not FAA ones).
It is possible though that they may ask a civilian contractor to have an A&P, but this would be on a contract-by-contract basis. The US Military does operate a number of civilian-type aircraft that they may have serviced.

Active duty military, no.
Civilian contractor, yes.


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Are there required standards for gate power cable maintenance?
I am looking for reference(s) addressing gate power cable maintenance, for both 28V DC and 400Hz. Do the FAA or any manufacturers have standards for measuring the connector pins to determine wear?

What would be the purpose of such a measurement? Is it a safety concern?

An undersized connector pin can cause electrical impulses on the aircraft structure. This in turn can damage LRUs. In extreme cases, it can place 115 Volts, 400-Hz on the structure. The same type of pin wear can occur on the aircraft ground power connector.

To answer your direct question, yes, I believe it's a major concern. Electrical noise can degrade Analog-yo-Digital (A/D) circuits used in sensors to convert analog signals to digital words the flight control system can process.

DC LRUs aren't going to be affected by bus power noise, as their input voltage is typically designed to handle 10-32 VDC. Any sensors are going to have their own power regulation as well, so I don't think that's a concern either.

Sorry, but very respectfully, you're wrong. All LRUs are bonded by the mounting trays, to the structure. Within every WRA, almost all ICs are grounded to the LRU chassis. Consequently any noise voltage felt on the fuselage will be felt on the ground plane of most ICs. Even very low (<5 VAC) are known to cause what is known as "electron migration" on the IC substrate.


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For FAA currency purposes, must flight time be in an N-registered aircraft?
I currently hold a Commercial Pilot Certificate issued February,2018. My Flight Review has to be done no later than February 2020.
I have convalidated my FAA License to Colombia. The 90 day currency for carrying passengers applies to both FAA and Colombia.
If I fly Colombian airplanes, does that flight time apply to the 90 day requirement in the U.S.?
In other words, can FAA certificate currency be demonstrated in non-N-registered airplanes?

14 CFR 61.57(a)(ii) details "category, class, and type" but does not mention registration of the aircraft.

I’d say you would have to ask the Columbian authorities about that. they would be the final arbiters in that case as Columbia is a sovereign nation which operates independently of US laws and regulations.  Unless there is some ICAO standard for pilot currency which both governments honor.

Why would the Colombian authority be the final arbiters for US laws?

No they would be the final arbiters over Colombian laws and not US laws.

OP is asking about currency in the US. He doesn't have an issue with currency in Colombia

Oh I see - he is talking about operating an airplane registered with the government of Columbia in the United States.  The phrasing he uses to describe that is pretty nebulous.  It sounded more like he wanted to fly airplanes in Columbia.


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Logging time towards commercial pilots license
I am in a part 141 program trying to finish up my commercial pilots license. The airport we operate out of will be closed for a few weeks so I was wondering if I'd be able to rent a plane close to where I'm from to knock out some of the solo hours I have left. It wouldn't be a school plane and my instructor wouldn't be with me. Is this possible?

Do you have a PPL?

Your question is not very clear. If you are asking whether solo time logged outside the 141 program can count toward completion of the 141 program, I believe the answer is no. If you are asking whether you could finish your license requirements and earn your license under §61, the answer is yes.

Sorry for the lack of information but you answered my question, thanks!

I am assuming that you are already a private pilot working on your commercial ticket. There is no requirement to do your solo hours at one flight school. Just make sure your primary instructor knows what you are doing and is OK with it.


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Why don't some spoilers plates fully open in most airliners?
The following image illustrates my question:

What's behind the manufacturer decision of opening some spoiler plates less or more than the others? I noticed this is quite common in airliners.

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Would increasing the number of wheels of a Jumbo reduce the braking distance?
If weight was not an issue, would increasing the number of main gear wheels on a jumbo such as the Airbus A380 from 20 to 40 wheels, each equipped with braking disks and braking equipment, considering their extra weight, reduce braking distance.
Would it be any different if the current set up was used but with longer wheelstuds and each wheel was paired with an extra wheel, as with most semi-truck's (articulated lorry's) rear wheels?

Some aircraft use a drogue parachute to shorten their stopping distance, in case you weren't aware of that option.

No disrespect to GdD but Sylvian's answer is much better, you know you can change the accepted answer right?

Aren't thrust reversers just as -- or more -- important than brakes?  More wheels add more weight, complexity and stuff to maintain.  You should instead ask whether bigger reversers would reduce landing distances.

@RonJohn sure it's obvious more reverse thrust reduces landing distance? That doesn't always make it the best option though.

@Notts90 my question was "more important than brakes?", not just whether or not they're important.

@RonJohn that’s a separate question entirely.

@Notts90 "that’s a separate question entirely" Which I found the answer to here: https://www.airspacemag.com/flight-today/how-things-work-stopping-the-a380-27549065/

@RonJohn that's quite a good article and you could probably make a good self answer Q&A on the topic using that as a source.

Increasing the number of wheels with brakes will decrease the braking distance as there's more sources of friction to convert the kinetic energy. Doubling the number of wheels per brake (I think what you're asking in the second part of the question) wouldn't automatically decrease braking distance as the brakes are limited to absorb a certain amount of energy, but you could put a more powerful brake on a pair of wheels than a single wheel of the same size as there's more grip.
In reality this wouldn't be done because it's not required, runway length is determined as much by takeoff length than braking distance, you could halve the braking distance of all airplanes and you wouldn't be able to reduce runway length.

Isn't required takeoff length also affected by braking distance, though? You have to have enough runway left to stop (ideally without blowing tires or setting the brakes on fire) in an RTO.

@reirab RTO>?..

@Cloud Sorry. RTO = Rejected Take Off (when you abort a takeoff run.)

@reirab Huh? There's a reason why decision speed and rotation speed are different...

@MichaelKjörling Indeed. But there are also reasons why we try to keep them as close as possible. Additionally, if you do have a large gap from V1 to Vr, this will also increase your required takeoff runway length, as you'll need to have sufficient runway remaining after V1 to accelerate from V1 to Vr and climb out safely with an inop engine.

Actually there are two ruling factors. Brake capacity, and adhesion.  Increasing the number of wheels won't help adhesion because adhesion is proportional to weight on rubber. Double the rubber surface, half the weight per cubic inch,  nets out to same adhesion.   What would help is manipulating airflow to push down on main gear, giving more weight-on-gear than the aircraft weighs.  That's exactly what an IndyCar spoiler does.

Is there any reason why landing runways need to be the same ones as takeoff runways?

@Yakk Because there is only one runway?

Typically, a little gain is obtained.
Larger airplanes use anti-skid technology. Anti-skid works by modulating brake pressure to ensure the tires never skid. It's important to understand the relation between a tire's load, brake pressure, and actual retarding force. First, look at this image :
It shows how as you increase pressure and the wheel starts slipping a bit (turning at a rate slightly slower than when free rolling), the friction coefficient rises sharply. Then after the 10% slippage, the peak coefficient is reached and thus the retardation force reduces. Anti-skid aims to keep the slip raitio as close to 10% as posssible. Therefore the retardation force is maximized.
The brakes of airplanes are strong enough to fully lock the tyres, but this is detrimental (results in 100% Slippage, so maybe 20% retardation force loss). So they can and do maintain optimal slip, therefore using the entire available 'grip'.
Now what happens to this picture if you add wheels?

If the added wheel is not braked (like the nose wheels are) then you lose braking. Indeed, the braked wheels will have a reduced vertical load, so they will provide less retardation force. It's quite useless.
If we add braked wheels then we see the tyiniest bit of gain. Let's see why.

First look at this picture of tire load sensitivity (this one shows lateral force while cornering, but longitudinal force while braking exhibits the same features):
What is important to note, is that as a tire gets more loaded, its capacity to generate friction decreases.
Now for a jumbo: When 20 wheels share 500 tons of load, each tire sees 25t. with a default friction coefficient of 0.8, each tire generates 20t retarding force, so total of 400t retarding force.
Let's add 2 wheels in there. Now each wheels sees 22.7t of load. Their friction coefficient could now rise to 0.82, therefore providing 18.6t, for a total of 410 tons. Certainly far of the 40t that could be naively expected from adding 2 braked wheels! Only the load sensitivity nets us a gain.
It then becomes the usual game of trade-offs. How much mass is spent on bigger landing gear and wheels and additional brakes, versus the gain of harder braking (and thus even stronger landing gear legs?)
Apparently the Airbus guys decided the trade stopped at 20 wheels (because 25t of load on a single tire is the maximum the ground could handle) but they didn't even bother to brake them all, if I remember well only 4 of the 6 wheels of the main boggies are braked! So if you wanted to improve brake efficiency:

Install better tires on braked wheels
Brake the non-braked main wheels
Make some smart aero changes to increase aero downforce & drag
Add braked wheels (and prepare to house them in the main body... good luck!)
Brake the nose wheels (but that would be terribly hard because it was never designed for that)

There's an important effect that doesn't show up in your charts: maximum braking effort of an airliner is limited not by friction, but by heat generation.  An airliner braking as hard as possible is right on the edge of blowing tires and/or catching the brakes on fire; more brakes or more braked wheels will let it stop faster by spreading the heat over more mass.

@Mark Or, sometimes, they're a little past the edge...

That's a very elaborate proof of a simpler concept: adhesion is proportional to weight on rubber, not rubber surface area.  Increasing surface area decreases weight per surface area, which decreases adhesion per surface area in proportion leaving you right where you started.   But you needn't add more gear; just let some air out of the tyres.  If you have 10,000 lb on a 100psi tire, your contact patch will be 100 sq.in. Deflate to 80 psi and your contact patch is 125 sq.in.

@Harper, despite what you learned in school, adhesion is not proportional to weight on rubber, it's only approximately so.  The actual mechanics of friction are so complicated that the greatly simplified "friction = normal force * coefficient" is what gets taught at the undergraduate level and below.

Sure, I admit that's the rough version of the math, the high school version, and there's some optimizing possible.   This for sure: what you say about braking, and how the brakes are able to ride at 10% slip, pretty much shows adhesion is the limiting factor, not braking.  Now how to get more adhesion?

Here’s something on Physics.SE that you may find interesting.
It’s all about the force and mass, not the contact area.
More wheels, same brakes = no change
Adding more wheels but keeping the same number of brakes would have negligible affect on stopping distance, as the total force going through the wheels and brakes would be the same and the mass change small.
There would be more wheels but each one takes a smaller load so there’s no additional friction between the wheels and the ground. There maybe some additional forces from the extra bearings and some extra mass from the wheels but that would be negligible in terms of stopping distance.
There’s the same number of brakes so they are providing the same force so they make no difference.
Same wheels, more brakes = shorter distance (possibly)
If we add extra brakes to each wheel (or bigger brakes) that allows us to apply more force to the discs, then we have more braking force and therefore the aircraft will stop in a shorter distance.
There are some issues with this though:

Heat - brakes generate a lot of heat that needs to be dissipated, more/bigger brakes will generate more heat. If the brakes get too hot they will fail and then you may not stop in time.
Structural limits - putting more force through the structure may cause it to fail which could be catastrophic.
Wheels - put to much force/heat through the wheels and they may fail as well. That maybe bursting if the heat/forces are to much or simply skidding if too much force is applied too quick. There are similar systems to the ABS on cars for aircraft the can help with this (Thanks to Michael for the link).

More wheels, more brakes = shorter distance
Adding more wheels with more brakes solves the main issues with the “same wheels, more brakes” scenario.
Why don’t aircraft manufacturers do this? Well all them additional wheels and brakes are mass that’s expensive to transport unnecessarily.

you missed something- if the brakes are already stronger than the tires ability to stop, for example like most passenger cars in low speeds- you can break hard enough to make the car skid, than adding more wheels (or just better tires) can potentially shorten braking distance

@Rsf Or use better brakes which don't cause the wheels to lock as easily. Those aren't found just on cars; airplanes have them, too.

@Rsf I was counting that as failure of the wheels, I’ll be more specific.

@MichaelKjörling: Anti-lock brakes work by decreasing the braking force if it would otherwise exceed the available wheel-road frictions. So "better brakes" will not allow you to overcome that limitation.

@HenningMakholm antilock brakes don’t simply limit the force. They constantly adjust the force based on current rotational acceleration, so you don’t go over the maximum force the wheels can deal with.

@Rsf: But more wheels won’t increase friction.

@Notts90: But that does effectively limit the force to what the wheels can deal with.

@HenningMakholm that’s exactly what it does yes.

Nope, and it's because of how tires work.
The limits are tire adhesion and braking capacity.  The latter is not the limiting factor - adhesion is.
Adhesion is roughly proportional to weight on tire.  Specifically, contact patch area (sq.in.)  x weight per square inch, but you notice square inches factor out, leaving only weight on tires.  It's more complicated than this, this is the prep school version, and there is some tuning to be done -- but it's been done.  And it doesn't help that much.
Take my car (please).  It has 600 pounds on a front tire. How big is the contact patch? That's actually easy: the tire pressure is 30 PSI. 600lb divided by 30 lb/sq.in. = 20 sq.in.   Contact pressure is, you guessed it, 30 pounds per square inch.
What happens if I reduce my tire pressure to 20 PSI?  OK, now 600/20 = 30 square inches of patch area.  With 20 PSI contact area.
Which brakes better? It's pretty close to a wash.  30x20 ~= 20x30 =~ 600.
What if I add lead ballast to my front end?  Now my weight is 750 lbs/tire (25% more).  YES! I get 25% better adhesion and thus braking power.  Except I now have 25% more mass to stop! So it doesn't slow any more swiftly.

OK, so you take a 747 and replace its 4-wheel bogeys with magic 6-wheel bogeys that weigh the same.  50% more wheels, right?  50% more contact area, right?  Nope. Think about how contact area is determined.  Same tire pressure but each tire is carrying 2/3 the weight as before, so the contact patch is 2/3 the size. You end up with exactly the same contact patch (in square inches) and exactly the same pounds per square inch as you had with the 4-wheel bogeys.
Anyway, if you wanted to increase the contact patch 50%, you just reduce the tire pressure 33%.  But just like my car, that wouldn't help you either.
All these factors wash out (roughly).  The ruling factor in adhesion is the weight on bogeys.
If only there was a way to increase weight on wheels without increasing vehicle mass.

There's the answer.  It's an aerodynamic lift surface that pushes down on the tires.  those cars go plenty fast enough for the Marlboro to be aerodynamically effective.  This works.  Weight-on-tires is increased so adhesion increases in proportion. Vehicle mass is not increased so it brakes correspondingly better.
Do the same thing on an airplane.  Come up with a way to attach a big wing of some kind directly over the main gear.  And set it up to generate downforce.
Not sure how to do it, though.  Maybe this?  ;-)

The lift dumpers do help.

Also, at high energies (where deceleration is limited by wheelbrake energy dissipation rather than by adhesion), adding more brakes does increase the available braking force.

It's a tradeoff, like just about anything involving aircraft design.
More wheels + more brakes means a shorter stopping distance. But...
More wheels means more weight, quite a bit more, to accommodate the extra landing gear, extra retraction/extension gear, extra storage while retracted, extra maintenance for all that extra stuff, etc... that means fewer paying customers and higher operating costs. This also means more things that can go wrong.
More braking means more loads subjected on the paying customer. I recall being on a C130 when the pilots practiced their short field landing: full reverse props just before touchdown, full braking. We hit (and I do mean HIT) the runway with a very loud clang, then we were slammed forward as the plane came to a very quick stop. It was... dramatic... fun when you're a twenty something in military training, but I'm not sure paying customers would like that.
As others have mentioned, the limitation is also takeoff distance. Doesn't do a lot of good to be able to stop in 500 feet, if the takeoff distance is 2000 feet. You'l still need a long runway if you want to get the plane out of the airport.
The one possibility would be safety in the case of an emergency, but the number of aircraft accidents that could have been prevented with more wheels for braking is extremely small.
Current airliner brakes are designed to meet the requirements, while remaining as light and as reliable as possible.

Absolutely, properly designed, more wheels and more brakes will get you a shorter stopping distance.
However, there is a lot more goes into the design of the landing gear and braking system than you might imagine. Braking distance, and force, is just one factor that is considered as part of that design and is more or less fixed for a particular aircraft style since runway lengths are defined by take-off more than landing.
Other factors which are balanced to achieve that obviously include weight, space lost to stow the landing gear, cost of manufacture, cost of maintenance, reliability and importantly, heat dissipation. Stopping a wide body jet generates A LOT of heat in the brakes which needs to be dissipated quickly to prevent overheating and fire.
Doubling up wheels can, with added stronger brakes, get you more braking, but you also now have to have a stronger carriage and fuselage attachments to absorb that braking force, and again, a larger wheel bay.
How much you want to brake is also dictated by passenger comfort. Brake too hard and you are open to passenger injuries and lawsuits.
That is a complex set of conflicting requirements. Adding more wheels and brakes would make stopping easier, but at a significant cost.

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text:
What is the air temperature inside of a tire right after landing?
What is the maximum design air temperature inside of a tyre of an Airbus A380's and a Boeing 747-800's main gear fitted with brakes, before landing after a 10 hour flight and right after landing?  Are there instruments to measure this? Do pilots take readings before and after landing?
Are there instruments to warn pilots about tire irregularities prior to landing? If so, what is the procedure then?

If your 2nd paragraph is meant to ask about warning pilots about irregularities prior to landing is asking about irregularities in general that affect the landing, I suggest asking that as a separate question. If it's meant to ask about tire irregularities only, I suggest specifying that in the paragraph.

Airplanes have tire pressure and brake temperature monitoring systems. Both Boeing (PDF and PDF) and Airbus have those systems installed.

I'm not sure that there is a maximum design temperature and even if there was it wouldn't be that high, probably no more than 120C.
The problem is not the temperature inside the tyre but actually the temperature internal to the carcass. The hot spots are around the bead area and around the cords on the shoulder of the tyre due to the loading. The rubber compounds will start degrading above 160C. This will be considered in the design of the tyre.
The design will have been tested under representative speeds and loads on a rig and the temperatures checked to make sure that the temperatures are within the required limits. At the Goodyear Technical Centre they have a large steel flywheel which they rotate to represent landing speeds and then use a hydraulic ram to simulate the landing gear and push the tyre onto the flywheel. They cycle this several times and then measure the temperatures throughout the carcass.
An under inflated tyre will suffer increased deflection of the carcass and be prone to over heating. Hence the main concern is then to maintain the correct operating pressure. The tyre will have charts specifying the correct pressure to use for the loading and landing speed.


text:
Is it possible to build a honeycombed vacuum balloon?
Is it possible to build a vacuum balloon with a near perfect void that can lift heavy objects and doesn't collapse and buckle under the atmosphere?
I expect perhaps a honeycombed or lorimerlite lattice infrastructure with toothpick-like filaments spanning across the honeycomb or lorimerlite voids in an alternating double helix pattern. These "toothpicks" would be slightly thicker in the center and thinner at their ends where they connect to the inside of the honeycombed or lorimerlite lattice. This would be made made of titanium. I would choose a toroidal (donut) balloon.
Would this design work, or is there another design that would?

I seriously doubt the titanium structure could achieve average density lower than hydrogen.

Using vacuum doesn't get you more than a marginal advantage over hydrogen.  Air averages ~29 g/mol, H2 is 2 g/mol.  The 0 of vacuum is only about a 7% improvement over hydrogen, and 14% over helium.

Air pressure at ground level is 1 kg/cm2. That's how much force your vacuum tank has to withstand.

There is no material strong enough to contain a vacuum that doesn't weigh more than the displaced air plus equivalent aluminium structure.

@jamesqf Of course, a vacuum doesn't explode.)

Vacuum would implode instead?

@Pilothead: Helium doesn't explode.  Nor does hydrogen, if you keep it well-separated from air.  And even then, it doesn't explode unless very well mixed, it just burns.

@jamesqf I think the hydrogen distinction would be lost on the passengers of the Hindenburg. I agree with your vacuum vs light gas argument.

You can load a decent titanium alloy to about $500 \mathrm{MPa}$ (the ultimate compression strength is quoted at least $850 MPa$, but you need to stay below the elastic limit and have some safety margin). That is $5000$ times more than the pressure at sea level. Therefore, the supporting elements have to take about $\frac{1}{5000}$ of the volume. At density about $4500 \frac{\mathrm{kg}}{\mathrm{m}^3}$, the average density comes at least $0.9 \frac{\mathrm{kg}}{\mathrm{m}^3}$. That is just slightly better than the $1.225 \frac{\mathrm{kg}}{\mathrm{m}^3}$ of air, but way worse than the $0.09 \frac{\mathrm{kg}}{\mathrm{m}^3}$ of hydrogen or the $~0.17 \frac{\mathrm{kg}}{\mathrm{m}^3}$ or helium. And this is only weight of the support—for containing vacuum, the skin will be much heavier as well.
Therefore there is no way this could be better than traditional hydrogen or helium-filled aerostats and it's unlikely to get off the ground at all in practice though the above back-of-the-envelope estimate suggests it just barely might.

Note: that calculation is hyper-optimistic. The best-case crushing strength of 3D materials is CS(d/D)^1.5, where S is original material strength, D its density, d the foam's density, and C<1. This limit is mathematically impossible to exceed with any structure and any material. At d/D<100 with any known metal, the collapse is elastic at C times S*(d/D)^2. So average density to withstand air pressure will be ~10 kg/m3 ideal (C=1, 1 GPa loading, plastic collapse) and ~50 kg/m3 realistic (elastic collapse).

If there was a reason to do this, ideally, you would want to use a structure similar to an open cell foam with a focus on minimizing cell size. Metallic microlattices would do much better than honeycombs.
But even better than microlattices would be some very compressible structure that would automatically adapt to the outside pressure. You need a very strong structure to withstand atmospheric pressure at sea level, but dense air also provides a lot of lift. As altitude increases, aerostatic lift decreases. So you need an increasingly lighter structure, but it will be under increasingly less pressure and can be much weaker.
A material that solves this problem actually exists - it's called gas. As atmospheric pressure decreases with altitude, gas expands and makes itself less dense. With the gas vented outside, the balloon literally makes itself lighter as it ascents.
The problem with the 'vacuum balloon' concept isn't even the lack of sufficiently strong materials. It's the decrease in lift with altitude. A balloon 100 meters in diameter will produces 5,000 tons of lift at sea level, but less than 2.5 kg of lift at the Karman line, the lower edge of space.
In theory, if the envelope could be light enough, an edge of space balloon isn't absolutely impossible. But, for any given finite level of material strength, it will fly higher with hydrogen (or helium) inside than with vacuum and a structure.
Any hypothetical material is better used to make a thinner shell to contain more gas. Since there's literally no limit to how light a gas can get, as pressure is reduced, there's no altitude where you'd have a reason to switch to vacuum.
There's also a practical advantage for gas - it equalizes the pressure, so a gas-filled balloon won't suffer explosive recompression and rapid sinking once a tiny dust speck makes a hole in it.


text:
Could a 'wind wall' help to reduce the wind shear problem at St Helena airport?
I've seen a fair number of articles online about "The World's Most Useless Airport" (St. Helena, HLE/FHSH), which was unable to open to full commercial flights (as was the intention) due to the wind shear that occurs there. Because of this, only small aircraft are allowed to operate into St Helena.
This may well be a completely impossible idea, but if the issue is wind shear could they not adopt a system similar to this 'wind wall' which they use to protect large cargo ships in canals in the Netherlands?
Obviously we are talking about aircraft here and not ships, but would an approach like this - diffusing the wind either side of the runway threshold - have any effect on the final stage of the approach, to the extent that the wind sheer becomes tolerable for 'standard' commercial air traffic? Obviously large concrete structures wouldn't be particularly sightly, but from a purely practical point of view would something like this alleviate the issue at all?

This is an interesting idea, but I think it might not be feasible as aircraft are much more sensitive to wind sheer than ships . If a ship is hit by a strong wind sheer it will drift to the side, if an aircraft is hit by a strong wind sheer it coulld end up side down.

Yes, I am sceptical anyway. I can't find very much information on why there is such bad wind shear for southbound landings and tolerable conditions for northbound landings, but  one source seems to suggest it is the fault of the small mountain on the edge of the plateau at the northern end of the runway. Maybe some dynamite could solve the issue rather than concrete :p

Who do you think you are, Donald Trump?

@BenTurner: Looking at the pictures, I would suspect part of the problem is downwind in southern direction. There's just nothing there, a steep drop-off in 3 directions. If you've got wind blowing from the southwest, southeast or straight from the south, it will "pop up" over the runway. A construction on those slopes might help, but I'm not sure if it needs to be a wall.

It's a completely different problem, and I'm afraid something like that would make the wind shear close to the ground actually worse since it is inhibiting air flow within 200 ft of the ground while the wind just above is unaffected.
Plus the additional turbulence created by those thingamajigs sticking up.  Most pilots know how unpleasant it can be trying to land on a runway with a stand of tall trees adjacent to it, with the wind shadow and burble generated by the trees.
The windshear problem for aircraft isn't the wind velocity itself, it's the sudden changes in horizontal velocity at certain levels, which in a large aircraft with a lot of mass, creates large airspeed fluctuations as you climb or descend through, along with vertical flows caused by mountainous terrain can can generate high sink rates that, because of the mass, are slow to be addressed with power.  None of that would be helped by a surface barrier.   All the barrier would do is create an extra nasty shear + turbulence right at 200 ft.


text:
Is a ground station radio license required for UAS operations in the US?
When conducting UAS operations would a PIC or Spotter be considered a ground station or an air station?  The communications would concern a sUAS in the air, but the radio is physically on the ground.
I'm asking in the context of licensing within the United States.  As far as I can tell from reading the FCC's website, Air stations do not require a station license to operate, while Ground stations do.

Welcome to aviation.SE! You might get a better answer if you tell us why you need to know, e.g. is it in the context of needing a radio station license? And if you're asking about regulations or other rules, please always tell us which country you're asking about.

@Pondlife This is in the context of needing a license within the United States.

This question seems poorly thought out, and practically moot.  If you were actually doing this, you'd be navigating the more complex regulatory landscape of flying it, and of licensing your control channels.  If you are just imagining it, consider that a ground-based transmitter for proxy communication from the UAS to other aircraft, ATC, airports, etc isn't going to work well - you'd need to transmit from the UAS to have decent radio propagation to these.

Although I can't pinpoint a FAR or FCC reg about it, an suas pilot should not be transmitting on air band frequencies except in emergencies. Receiving never requires a license in the us

@TomMcW the sUAS training materials from the FAA recommends you listen to aircraft frequencies and does not recommend transmitting, but does not state it is prohibited from the FAA's standpoint.  The FAA does not cover other regulatory agency requirements that intersect with sUAS flight, like NTSB reporting. I would assume they are ignoring any FCC requirement as well. Admittedly, the question is slightly academic due to practical limits of line of sight communication, but still useful to have answered.

@ReidCrowe I think people are overestimating the limits of a handheld. Instructors talk to students on solo flights often. ATC antennas are on masts, so if you're anywhere near a tower I'm sure they could hear your handheld just fine.

According to the FCC website on aviation radios

On October 25, 1996, the FCC released a Report and Order in WT Docket No. 96-82 eliminating the individual licensing requirement for all aircraft, including scheduled air carriers, air taxis and general aviation aircraft operating domestically. This means that you do not need a license to operate a two-way VHF radio, radar, or emergency locator transmitter (ELT) aboard aircraft operating domestically. All other aircraft radio stations must be licensed by the FCC either individually or by fleet. [emphasis mine]

There is no specific mention of UAS, so I would say that you could not use it from the ground. It also specifies the following with regards to handheld radios:

You may only use your hand-held aircraft VHF radio in your aircraft under the terms of your aircraft license. You are required to have a separate Ground Station license to operate your radio on the ground. [emphasis mine]

It's fairly common practice for flight instructors to talk to students on solo flights via handheld, but it was decided in a discussion (can't find the question) that one technically is supposed to be in a plane to do so.

I would be interested in reading the discussion you are referring to.  Please post a link if you come across it.

@MichaelHall I still can't find that discussion. I think it was in a comment thread somewhere, but it may have been in chat. Either way, it was just an informal discussion.

Air -> Air, or Ground -> Air describes the ultimate path of the message. It doesn't describe what's being discussed. So:

The communications would concern a sUAS in the air, but the radio is physically on the ground.

If the radio endpoints are both on the ground, then that's ground -> ground transmission. Put it another way - if a controller talks to a fireman over the radio, about an aircraft in the air, it doesn't magically become a Ground -> Air transmissions.
The definition matters primarily for technical and practical reasons (E.g.., Air -> Air radio will have a better range than Ground -> Ground because of nothing being in the way). If you're on the ground, you're a ground station - it's as simple as that.

Definition also matters for licensing purposes.  As I understand from the FCC's website, Air stations do not require an FCC license while Ground stations do.
Edit due to premature posting:
The context of this is if a remote PIC needs to communicate with Aircraft about the location of the sUAS, from a licensing standpoint is the remote PIC an air station or ground station.


text:
Why is the Aerodynamic Centre not affected by changes in Angle of Attack?
I understand that the Aerodynamic Centre (AC) is the point in the aircraft where the pitching moment never changes with a change in Angle of Attack (AoA). My understanding of the Centre of Pressure (CoP) is that it is the resultant pressure point in the airfoil and it slides rearward when AoA decreases and vice-versa, i.e. it changes its position with regard to the AoA.
Now the point is, how does the pitching moment (the force arm, i.e. the distance between the CoP and the AC multiplied by the force) remain constant when the CoP itself moves with the change in AoA?

Welcome to aviation.SE! I've edited your question to try to make it clearer. If I changed it too much or made a mistake, please roll back or edit again. And you might be interested in the [tour] if you're new to this site.

As the pitch of the wing increases moment arm decreases and at the same time lift increase, so the pitching moment stays constant.
In typical airfoils such as the wings of an airplane flying level sub-sonically the pressure center is at approximately 0.44 of the  cord length, and the aerodynamic center is at 0.25 of the cord length.
When the AoA increases the lift increases linearly and at the same time the CoP moves forward closer to aerodynamic center in a way that reduction in the length of the pitch moment arm is cancelled by increase in the force (lift); so the pitch moment is constant.
In fact the location of AC is calculated by finding the point that the decrease in pitching arm is proportional to increase of lift, otherwise one could pick any arbitrary point along the cord and calculate the pitching moment at that point, but it would not remain constant for a different AoA.


text:
Why does fuel vaporize in the carburetor?
So I get that a carb vaporizes fuel and mixes it with the air intake for the engine. But my question is what is the explanation for what causes the fuel to vaporize after it exits the fuel nozzle?

This is a great video that explains how carburetors work and shows in slow motion exactly what's happening inside: https://www.youtube.com/watch?v=toVfvRhWbj8

Vaporization is closely related to the concept of volatility, which in turn is directly tied to the vapour pressure of a substance.
The vapour pressure of a substance is a function of temperature, and is the pressure at which a room filled with some liquid (or solid) and some gaseous parts of the substance are in exact equilibrium. Increase the pressure, and some gas will want to condense; decrease the pressure and some liquid will want to evaporate. For fuels, this is in the order of tens of kPa (e.g., the Reid vapour pressure of avgas, which is the vapour pressure at 37,8°C, is ~38kPa).
If we assume that air is an ideal gas, we can say that liquid fuel and gaseous (evaporated) fuel are in equilibrium when the partial pressure of liquid fuel equals the vapour pressure of the fuel. Since the partial pressure of the constituents of a gas are directly tied to the molar quantities of a gas, we can easily extract the fuel-to-air ratio. Given that the partial pressure of oxygen is about 21kPa (which will only decrease if you add fuel, assuming this happens with constant atmospheric pressure), we can easily see that if you leave a tank of air and fuel (with a vent to keep it at atmospheric pressure) long enough for the fuel to reach equilibrium, you have a very rich mixture indeed.
So, we limit the amount of fuel entering the carburetor to close to stoichiometric (exactly enough fuel for the amount of oxygen). As per above argument, all of this fuel will eventually evaporate as the vapour pressure is much higher than the eventual partial pressure reached by the gaseous fuel. The only question that remains now is how fast we can evaporate all this fuel. This only happens at the surface of the liquid fuel, so the trick here is to maximize this surface area. This is done by spraying the fuel into tiny droplets, each of which has a large surface area compared to its volume (remember, area scales with the size squared, while volume scales with size cubed - if you make the droplets smaller, the volume will decrease more rapidly than the surface area). They will thus be fully evaporated before they reach the actual cylinders. The smaller the droplets, the quicker this happens and the better mixed your fuel will be into the air.


text:
What is the difference between a forward slip and a side slip?
I am having a hard time understanding what differentiates the two types of slip. I understand that in a forward slip, the longitudinal axis is tilted with respect to the flight path, whereas in a side slip it is not. In a forward slip your track stays the same, whereas in a side slip it would (in the absence of wind) cause you to track toward the down wing.
My difficulty comes from the fact that the control inputs needed to enter both are the same: banking with opposite rudder.
Why then is the heading of the aircraft with respect to the flight path different, if the control inputs are the same? Is it just a question of degree, which is to say, the forward slip just requires more rudder?

Related : https://aviation.stackexchange.com/questions/91403/viable-to-slip-onto-the-center-line/91428#91428 -- the sentence in the main text that is associated with footnote 2, tells exactly what the difference is.  It's in how you enter the maneuver.  Should develop this into a full answer but not today--

There is no difference aerodynamically. The only difference is in intention and presence of the wind. The airplane does not care about the ground track, all it feels is the movement through the air. Both side- and forward-slip make the airplane fly slightly sideways through the air, somewhere in the direction between nose and downwards pointing wing.
If there is no wind, your ground track will be exactly the same, between wing and nose, and you will feel that the nose points away from your track and call it forward-slip. If the wind is blowing and you manage the right amount of slip, you are moving sideways through the air, but the ground track (combination of wind- and slip- movement) can be aligned exactly with the nose and you call it side-slip.
There is really no difference aerodynamically and thus no difference in applied controls either.

So it is a problem of reference frames more than anything. The important reference frame (the air) being invisible makes them look different when in fact they are the same thing. Thanks!

This is correct, and makes a clear distinction between airflow and wind. The aircraft encounters only airflow and has no concept of 'wind' - the only thing that knows of wind is you, seeing the ground go sideways.  The air through which you are flying is still.  The still air through which you fly, is being carried across the ground.  The latter situation is what wind causes.

Awesome answer! Thank you for being one of the few who understands the perceptual difference created by misunderstanding of the effect of which frame of reference we are focused on on our mental assumptions.

Sideslip and forward slip are actually the same thing.  They are both intentional cross controlled, unbalanced flight.  The only difference is where the runway is oriented with respect to the aircraft's attitude, and the reason for executing the slip.
In what is being called a "forward slip", (used to lose altitude without gaining airspeed) the winds are approximately straight down the runway, so the aircraft attitude must be cocked to the side to keep the flight path lined up with centerline as the aircraft actually sideslips at an angle towards landing.
In a crosswind situation the reverse is true:  The aircraft's flight path relative to the wind is into the direction the wind is coming from, (angle of bank into the wind to keep from drifting) but the aircraft is cross controlled with rudder to keep the nose lined up with centerline.
The aircraft's attitude with respect to the relative wind is the same in either case.  Any time an aircraft is in unbalanced flight it is slipping both forward and sideways to some extent.  The only difference is whether you are compensating for a crosswind or not, and where the runway is on your windscreen.

The difference is in application.  First, the forward slip (my favorite). The inputs (for a 172) are full rudder and opposite aileron as needed to hold straight path. This is cross controlled, but safe if you maintain a healthy margin above stall. I did mine at 65 knots.  The purpose is to increase drag, producing a faster rate of descent for a given airspeed.
A side slip is used to hold your track in a cross wind.  Here you aileron (drop wing) into wind as needed to stop being pushed sideways, using opposite rudder to control "weather vaning" of tail, and a tad more up elevator.  This is where you want to be just prior to touchdown.  Again cross controlled.
These are best practiced at altitude first, as different aircraft have different cross controlled stall characteristics.  The 172 produced some tail oscillation in a forward slip with flaps down, but, especially at 65 knots, got me home every time.

I understand the difference in application. What I don't understand is how the same control inputs can cause both situations. How can the same control inputs end up causing two very different attitudes?

They are not the same.  Think of what you are doing.  In a forward slip you skid the plane sideways for more drag, using opposite aileron to keep it from turning.  In a side slip you roll the plane into the wind to hold a ground track, and use opposite rudder to keep the nose on that track.  You will better understand it with some practice.  It was confusing to me at first as well.

It sounds to me like the main difference between the two situations you describe are the order in which you named the control inputs. Ignoring the external situation (wind), it sounds like they are the same practical thing, and what actually differentiates them is context. Isa this accurate?

@KBriggs sounds like you really want to compartmentalize this issue, I guess as "cross controlled" flight.  The only thing I can add is when you fly, you cannot ignore the external situation, in fact, you are a part of it.  Your control inputs are to make the plane do something, depending on the external situation and your flight plan. You side slip to hold ground track against a cross wind.  You forward slip to increase drag for a steeper rate of descent, a bit like adding flaps.  What confused me was they were both called "slips".

@KBriggs you are correct in that the control inputs are the same, and the resulting action of the aircraft is the same.  The only difference is the presence of wind, which affects only your ground track, and cannot affect the flight characteristics of the aircraft.  Refer to the answer of Martin above.

Forward slip and side slip are the same.  It's just the objective, degree of control input, and alignment of the nose is different.  Forward slips are to lose altitude.  Side slips are to compensate for lateral drift while landing.
In a forward slip your path through the air is more or less the same as before you started the slip.  In a side slip your path is skewed laterally (kind of a forward slip with insufficient input).
Whether it's forward or side slip depends on the amount of control inputs.  A forward slip has enough rudder to point the nose to the side, with opposite bank to prevent turning, so that your path continues straight ahead in the absence of a crosswind.
A side slip uses rudder to keep the nose aligned with the original heading, and with bank applied to let the airplane "slide downhill" so to speak, into wind, the result being the airplane tracks along the ground where the nose is pointed even though there is a crosswind.

Yes, forward and side slips are perfectly identical.  When used for landings the difference might just be a matter of which way you are looking out of the cockpit.  Let's say we are on short final with a crosswind from the right that is about half of what control authority will permit.  That is, rudder about half deflected to the left and right wing down to counter turning tendency, we are lined up with the runway, pointed along the runway, and not drifting left (or right either) just tracking right along the runway extension.  A nice stable side slip (with a steeper than normal glide path due to the cross control if not carrying additional power).  At that point we suddenly realize we are higher than we want to be so we smoothly apply a little more left rudder while smoothly increasing the right bank to add a bit of forward slip on top of the side slip to steepen the glide path.  If done well the airplane continues along the same track, only now we are looking slightly to the right to look at the runway.  Afterwards, having joined a more desirable approach path we decrease the rudder input back to where it was for the original side slip, while smoothly decreasing the bank angle to balance the reduced rudder and now we are once again looking right over our nose at the runway.  Continue to the flare and touch the right main, left main and then I surely hope the tailwheel!

Forward and Side-slips are not quite the same although they are very similar.
When a plane is put into a slip it tries to roll out of the slip. For example if I apply right rudder this puts me in a left slip (relative wind hits the airplane from the left) and the plane will try to roll right.  This is called slip-roll coupling.  Also the slip will try and turn the plane in what John Denker calls a "boat turn" in his book.
In a forward slip, the turning tendency by the slip ("boat turn") is exactly counteracted by the bank of the wings (what John Denker calls a "momentum turn") so that the airplane does not turn.  This means that the ground track remains the same but that the airplane will be in a high drag / low lift situation that allows the plane to lose altitude without gaining airspeed.
Side-Slips are similar except that the turning tendency of the wings and resulting slip are not equal so that the airplane will try to maintain a constant heading while in a slip.  This is useful in a cross-wind landing where the cross-wind will try to blow the plane off center line if you not crabbed and trying to keep the nose aligned with the center line.  The correct for the crosswind the airplane has to turn into the wind while keeping the nose aligned with the center line.  This is accomplished by having a larger bank ("momentum turn") than degree of slip ("boat turn") so a little less rudder is applied in a side-slip than in a forward slip.  Side-slips are usually defined as maintaining a constant heading during the duration of the slip.
In both instances you are cross controlled its just that in the slide-slip you will use a little less rudder.
Here is a graphic from ClayAviation showing the concept:

Tilt the entire diagram above, (or your head), and you will see that they are identical!


text:
What's the minimum legal height in Europe for a low pass at an airport?
Does anyone know in which document I can find the legal minimum height for a low pass at an airport, assuming the pass has been approved by ATC? I have been discussing this topic a lot with several flight instructors but we haven't found any official document where a specified height is set.
I'd like to know about EASA regulations if possible, but FAA or others are also appreciated.

What do you mean by low pass @Flightfighter? An aborted landing is a low pass, so if all you want to do is fly low over a runway then all you need is clearance at a towered airport. At a non-controlled airfield you can just do it.

You are right, an aborted landing is a low pass, but what mean is: when you are cleared for low pass at any aerodrome, because your intentions are not to land but only to fly a low pass, there is a minimum legal height, or is it allowed to fly as low above the RWY as you want? You know what I mean @GdD ?

I don’t think that’s regulated - never seen it, anyway! I think it’s go as low as you want, but if the gears touch, you pay the landing fee. Hard to give negative proof though...

At least in the US I've requested "the option" and flown in ground effect/flare (nearly) the entire 8000 foot runway. I was probably no more than 2-3 feet off the ground. I've also done the same with just one wheel on the runway.

Technically as low as you can without touching the runway. In AF training my instructor told me to "go around at beer bottle height", and is a legal low pass.

So I guess nobody has an answer for this right?;)

Official Record Series 4 No 1174 General Permissions
Standardised European Rules of the Air – Exceptions to the Minimum Height Requirements
1) Definition
In these permissions:
a) ‘SERA’ means the Annex to Commission Implementing Regulation (EU) No. 923/2012 (‘the Standardised European Rules of the Air’ (and references to SERA followed by a number mean the corresponding provision of SERA)).
b) ‘licensed aerodrome’ means an aerodrome licensed under the Air Navigation Order 2009 or an aerodrome certificated under Commission Regulation (EU) No.139/2014 of 12 February 2014 and
c) ‘Government aerodrome’ means any aerodrome in the United Kingdom which is in the occupation of any Government Department or visiting force (as defined in article 255(1) of the Air Navigation Order 2009).
d) ‘notified’ has the same meaning as in article 255(1) of the Air Navigation Order 2009.
2) General (SERA.5005(f)(2))
a) The Civil Aviation Authority (CAA) permits, under SERA.3105 and SERA.5005(f), subject to the condition set out in subparagraph (b), an aircraft to fly elsewhere than as specified in SERA.5005(f)(1) at a height of:
i) less than 150 metres (500 feet) above the ground or water; or
ii) less than 150 metres (500) above the highest obstacle within a radius of 150 metres (500 feet) from
the aircraft.
b) The aircraft must not be flown closer than 150 metres (500 feet) to any person, vessel, vehicle or structure except with the permission of the CAA.
3) Approaches to Landing or Forced Landings
The Civil Aviation Authority permits, under SERA.3105, SERA.5005(f) and SERA.5015(b), an aircraft to fly below the heights specified in SERA.5005(f) and SERA.5015(b) if it is flying in accordance with normal aviation practice and:
a) practising approaches to land at or checking navigational aids or procedures at an aerodrome;
b) practising approaches to forced landings other than at an aerodrome if it is not flown closer than 150 metres (500 feet) to any person, vessel, vehicle or structure; or
c) flying in accordance with a notified procedure or when specifically authorised by the CAA in accordance with SERA.5015

Can I suggest formatting to make this more easily readable? From first looks though, I don’t think it answers the question (as in: If I approach to land but then don’t touch down, how low am I allowed to skim the blacktop?).

I am no expert on the subject, but I have been looking for a legal document about low approach and I haven’t found anything to respect minimum height. Then, I can only assume that is regulated under the above document. Again, I am no expert on the subject, but I would say, as stated on 3.B “... not flown closer than 150 meters (500 feet)”, be the minimum height. Unless we specifically are cleared for a Lower height in accordance with 3.c “ flying in accordance with a notified procedure or when specifically authorized by CAA...”

I‘d say it’s a), because you are practising an approach to land. It doesn’t matter whether you touch down or not, you will still fly the approach as if to land. So therefore, no minimum height applies for that scenario. Of course, my interpretation only, obviously :)

According to this, could a low pass be considered a notified procedure? "c) flying in accordance with a notified procedure or when specifically authorised by the CAA in accordance with SERA.5015" . If yes, then there could actually be no minimum height specified.

In the US we would just ask for the option.  Or a touch and go.  If cleared for a touch and go they wont punish you if your wheels never actually touch!

In the US, I fly out of an airport with a few flight schools. Because of the student traffic, Touch-and-Gos and Stop-and-Gos are discouraged by ATC during busy times. Touch-and-Gos are discouraged by the flight schools on all student solos. Low Approaches are a common practice during busy times. Tracking down the runway is a skill building exercise for landing practice. When not busy, you can request for either a Low Approach or the Option and receive clearance. You can even request to land long and wheelie down the runway. There is no minimum altitude unless you circle or “go missed”.


text:
What's a stepdown fix in a non-precision approach procedure?
Could someone explain to me what's a stepdown fix in non-precision approaches?

From the FAA's Pilot/Controller Glossary:

STEPDOWN FIX − A fix permitting additional descent within a segment of an instrument approach procedure by identifying a point at
  which a controlling obstacle has been safely overflown.

In other words, a step down fix tells you "you've passed an obstacle, so you can now descend further".


text:
What's the function of the plate behind the wheels of the Pilatus Porter?
Anyone knows what's the plate behind the wheel of the Pilatus Porter for?

Looks like a sort of gravel kit.
Real gravel kits are used by jet powered aircraft to land on rough airfields.
In this case though, the engine is in front, so it is likely that it is only there to reduce the amount of dirt raised when landing/taking-off so to reduce the chances of damaging the aircraft.
From the official PC-6 Porter brochure via pc-6.com (PDF), they call this option a mudguard:

MUDGUARD

And according to a service bulletin (pilatus-aircraft.com; PDF) there were prone to fatigue:

Operators have reported failures of the mudguard supports and stone guard mounting yokes. The failures are caused by fatigue damage.

I would suspect that without those plates stones might hit the lowered flaps during landing rollout.

Someone told me it could be used as a mechanical brake when landing on an upwards muddy runway,  sticking the plates in the mud in order to stop the plane if it moves backwards ( due to the upslope).  Does it make sense?

@Flightfighter it could make sense, but I would not expect it to be below the center of the wheel when "at rest" like in the pic. look at the right wheel, you can see that the supporting arm has a bend to lower the plate over the ground, such a bend reduces the load it can carry if used as a brake. Plus, the structure does not look like it is made to carry any load at all.

Thank you buddies, it was kind of obvious but just wanted to be sure:)

Federico hit the nail on the head.  It’s a gravel kit and works similar to the mud flaps on trucks.  On rough airstrips, it prevents rocks, gravel and other FOD from striking the airframe and damaging it during the ground roll.

How does this add anything to the already existing answer?

Actually - Federico didn’t exactly hit the nail “squarely on it’s head” The main function of these devices is to spread the load, ie reduce the weight per square unit by increasing the ground contact patch of the tyres. Many bush airstrips has a very loose / soggy sub-soil, and I’ve seen first-hand that a heavy bush-plane can sink through to it’s belly while sitting still or moving slowly when operating in unimproved jungle-airstrips in Asia or Africa. Think “snow-shoes” in the Arctic or wheel-penetration skies, like the ones I flew in Antarctica. If the “ground contact patch” of the tyre itself, starts to break the surface, it will sink down until the device touches the surface, whereupon the load is spread out to much larger area, and the aircraft doesn’t break through.

If it is supposed to take some of the airplane weight, the skimpy tubing that fixes that plate to the landing gear should be a bit beefier, wouldn't you think? On the other hand, the plate is wider than a pure mudguard needs to be …

Do you have a source for this?


text:
Why is the relation of height and temperature different when calculating altitude?
I'm trying to grasp the essence of altitude calculations. One thing I can't figure out in my head is this:
In the International Standard Atmosphere (ISA) the temperature of the air decreases with the altitude, which is called the lapse rate. This is an approximate of 1.98 C (degrees Celsius) or an even more approximate 2 C per 1000 feet. This is usually described through dry adiabatic lapse which means simply that when you go up the pressure rises and the air expands thus through the ideal gas equation it makes sense for the temperature to drop as well.
However when calculating the density altitude we need to "correct" the pressure altitude with this formula:
DA = PA + (118.8 ft/C) x (OAT - ISA temperature)

(from https://en.wikipedia.org/wiki/Density_altitude#Calculation)
Here the height change per degree Celsius is only 118.8 feet (approx 120 ft).
My question is this: Why is the relation of (change rate) height and temperature different? With just "going up" it changes 2 C / 1000 feet (ie 500 ft/C) but when adjusting for temperature it changes 120 ft/C).
These both are simple linear equations but have different coefficients.

You're misinterpreting the calculations, 118.8 ft/C does not mean that the temperature changes 118.8 C per 1000ft.

GdD, that's not even what I'm claiming. The change in altitude is either 118.8 ft/C or 500 ft/C

Sorry @user541905, that's actually what I meant, it isn't saying altitude changes one degree in 118.8 ft/C. Also, keep in mind that that formula gives you an approximation.

Adiabatic Lapse Rate refers to properties of a packet of rising air which differ from the environmental Lapse Rate of the part of the atmosphere it is rising though. The difference between the two is useful for predicting clouding levels and formation of thunderstorms.

The equation is saying, that for every degree difference from standard temperature, the difference between pressure and density altitude grows by 118.8 ft. A completely different factor from the lapse rate.

I get this, but the question is why is the difference 118.8 ft and not 500 ft like it is if we approximate the temperature change at different altitudes? If there would be a totally linear relation between height and temperature it should be the same all the time? Or is this about the fact that talking about feet is essentially hiding the fact that we should be talking about pressure? I get the feeling that talking about feet is trying to make physics "easier" for aviators, whereas I would understand talking about pressure better. (Altimeter is essentially a barometer with a weird scale)

Or maybe the answer I would be looking for (thinking about your answer and looking at this graph https://en.wikipedia.org/wiki/Density_altitude#/media/File:Density_Altitude.png) is that pressure has a bigger effect on density than temperature (that is to say that in the differential equations defining density, the slope or coefficient of the pressure term is higher than the coefficient of the temperature term to make an approximation)

@user541905, It is effect of temperature, alone, on density, at given pressure. So it is independent of the effect of pressure.

Because it's two different things:

the lapse rate is how temperature goes down while you go up in the standard atmosphere
the 118 figure is how to adjust pressure altitude to obtain density altitude when temperature differs from ISA temperature.

Check this post to see how that 118 is derived.


text:
Is it possible to add "planets" to the gearbox of a Trent 1000?
The gearbox of a geared turbofan engine has 4 reducing gear small wheels being spun by a larger wheel or gear, simultaneously spinning the shaft that is attached to the fan.
That is 8 points of contact, two for each small wheel/gear meaning 8 the points where friction and subsequent heat would result.  I gather the use of 4 small wheels/gears instead of one is for reducing the load on the teeth at each given point.
Would increasing the smaller gear wheels (i.e. the "planets" ) from 4 to 8 or 12 or 16 for a gearbox designed for an engine the size of a Trent 1000 or more render the gear teeth less likely to break or fail?
With some reinforcement on the teeth, would it be possible to make a gearbox for this large engine?
As for the extra heat, cooling and rotating the oil through a cooler or along the length of the aircraft several times over, exposed to the  atmosphere at cruising altitude using a high pressure oil pump passing oil through a sealed gear box would take care of cooling.
With the ambient energy loses to heat would this render the engine a lot more fuel efficient and possible?

how do you increase the points of contact? increasing the number of teeth? but then you have teeth that are way small. then you have to increase the diameter of the gears, and you end up with an gearbox that's way too big.

The problem with gearbox cooling is not cooling the oil afterwards, but circulating enough oil in an efficient manner to cool the gearbox in the first place. Perhaps this can help you focus your question towards what you really want to ask.

Smaller teeth means it can't hold as much torque, so I think adding teeth will make it more likely to fail.

While you mention 4 satellites, the PW1100 has 5 ones. Epicyclic gearing is a crazy technology and efficient gearboxes are difficult to built due to the complex plays to catch in spite of wear if you want all teeth to transfer the load.

The fundamental question you seem to be asking is whether adding more planets to a planetary gear improves the load carrying capability of the gearbox.  The answer to that question is yes... to a point.
Take a look at a picture of the PW geared turbofan gearbox.  You'll note that there are five planets in there, and they are packed in pretty tightly.  There's really no room to add any more.  Although adding more planets would distribute the load such that each planet takes less load, it's impossible in this configuration because there is no space.  Let's say that you really really wanted to make it six planets instead of five. The only way to make enough room would be make the planets smaller. As you make the planets smaller, then either the ring gear has to get smaller too, or the sun gear has to get larger.  Either way, that is going to change the gear ratio (see formulas here). Specifically, the gear ratio is going to get closer to 1, which means the gearbox no longer does the job it was intended to do (namely reduce speed and increase torque).
So you can't have your cake and eat it too.  If you want 16 planets in a planetary gear, you can do it, but the gear ratio will be something like 1.02:1 which is not useful.  You might as well forget the gearbox and just do direct drive at that point.  If you need a decently high reduction ratio like 2:1, you can do it, but you will be limited on the number of planets that you can fit in.

The PW gearbox of the geared turbo fan was designed for an engine designed for an aircraft the size of an A320.  One would expect the gearbox of an engine desinged for an A380 to be larger, hence a large sun.  Reducing the size of planets and increase the size of the sun would result in a larger gearing ratio.  Reducing the size of the planets would increase the space to add more planets and increasing th size of the sun would increase space for more planets hence more smaller planets and a larger sun would increase gearing ratio. if planets were the size of the sun gearing ratio would be 1:1

“Specifically, the gear ratio is going to get closer to 1” – this is not true. In the limit $N_\mathrm{s} \approx N_\mathrm{r}$, $N_\mathrm{p} \ll N_\mathrm{s}$, the transmission ratio $\frac{\omega_\mathrm{c}}{\omega_\mathrm{s}}$ approaches $\frac12$. I.e., it is possible to have a planetary gearbox with arbitrarily many planet gears, while still getting a useful torque conversion. It just probably doesn't offer as good a tradeoff between power, weight, reliability and efficiency-gain as the 5-planets configuration does.

@leftaroundabout I stand corrected. Not sure what I was thinking when I wrote the answer but you are right.


text:
Do large Airliners use full thrust when reversers are deployed?
Do large and heavy Airliners, Jumbos and Super Jumbos wide bodies included use full thrust for braking when reversers are deployed?  What is the thrust setting?

related: https://aviation.stackexchange.com/q/3660/1467

The answers given below don’t really specify how much actual thrust the engine produces when “max reverse thrust” is selected. I have a hunch it is considerably less than full rated “forward” thrust.

Flight manual procedures and operators‘ policies notwithstanding, they would typically use full reverse thrust when the stopping performance requires so, and idle reverse thrust when not. The latter setting means less wear on the engines and lower noise. The former setting overrides these considerations when safety demands it.

Depends on...
* the manufacturer's recommendations,
* the operator's policies,
* the airport's procedures,
* the Captain's wishes.
A 747 landing in LHR will probably only use Reverse Idle, whereas landing on a classicly-Caribbean airport may go to Full Reverse.
And the A380 only has 2 reversers, so that may affect the policy on that aircraft.

Unless, of course, the Heathrow 747 is landing on an icy runway...

Thrust reverser works best on contaminated runways. On a relatively dry runway, full vs idle thrust reverse has negligible effect on braking. However, on icy runways, full trust reverse has a much greater effect on stopping distance. This information comes from the Airbus operational manuals.

One could quibble about "works best" vs "makes the most difference," but whichever phrasing is used, the basic point that TR's are most useful on contaminated runways is correct. It would be helpful to use that fact to reach an answer to the stated question, which is how TR's are used in practice.

This doesn't actually answer the title or body questions.


text:
What is the maximum temperature of engine bleed air for de-icing the wings?
What is the maximum temperature of the air bled from the engines of a Boeing 747-800 and an Airbus A380 to the wings for de-icing?  If the air is too hot, how is it cooled?

Related: How hot would pressurized air get if the air conditioning failed?

I dont know about the maximum temp but in B777 the Wing anti ice doesnt operate with the temp below 93 degree C. The air comes from the main pneumatic duct and initially cooled through Precooler(Heat exchanger) by using fan air. The temp and pressure for the particular wing is then further controlled by its respsctive WAI(Wing anti ice ) valve.

You should narrow your question down to a specific type or design. I could give you book figures for multiple aircraft that I fly.

This is based on aircraft type and manufacturer approved limitations.  Typically it’s limited by temperature or duration of use.  Compressor bleed air is tapped at around 450° F, hotter than a typical household oven.  These high temperatures can begin to significantly affect the yield strength of aluminum alloys typically used in aircraft construction, approaching the design factor said of safety and potentially resulting in a local structural compromise during high load factors, etc.


text:
What powers the air required for pneumatic brakes in aircraft?
What powers the air required for pneumatic brakes in large airliner aircraft? Is it bled from the engines, or do they have a separate motor and pump?

For a medium size turboprop, the F-27: two separate compressors, driven by the turboprop engines.
From this article

Fairchild’s F-27 is the first American-made plane to use a complete pneumatic system to operate landing gear, wheel brakes, nose wheel steering, propeller brakes, and passenger door operation. It also has an emergency air supply that will lower the landing gears and operate the brakes.
Compressed air for the system is supplied by two 2-cfm, 3300-psi compressors located in each of the engine nacelles. The compressors are driven through a gearbox by Rolls Royce Dart engines, which power the aircraft. They carry their own individual dehydration and filtration equipment to supply clean, dry air at a pressure of 3300 psi

Most present day airliners use the hydraulic system for the landing gear, including wheel braking. For instance on the B737:

The normal brake system and autobrakes are powered by hydraulic system B. If brake pressure drops below 1500psi, hydraulic system A automatically provides alternate brakes which are manual only (ie no autobrake) and the brake pressure returns to 3000psi.

The hydraulic systems are each powered by an Engine Driven Pump and an Electric Motor Driven Pump.

This makes me wonder if the F-27 sounds like a semi-truck when it's taxiing. :)

Yes it would hiss if you were close to the brakes while they were being used but without that squeak noise from those big bellows units.  This was something more popular with Russian designs, and I think the Fokker was the only western transport a/c to do that.  Lots of technical advantages, but the difficultly of sealing HP air, and controlling humidity, in a system where you are using it to do hard physical work is a nightmare from a reliability standpoint. Once the fire risk for hydraulics was mitigated by the switch to phosphate ester fluid, pneumatic actuation became much less attractive.

Well, yes, the Russians have been manufacturing the very reliable Vedeneyev M14P radial engine for a very long time, and to my knowledge pretty much all of them are air-start. Makes perfect sense. Not a good idea to rely on a battery in the middle of Siberia. :)

As long as you religiously monitor for air leaks and chase them down.  The cylinder will hold its charge for months... until a tiny leak starts and then you're screwed, and it's the old hand bombing start method.  Years ago I was planning to get checked out in an air-start Wilga to tow gliders until I found out I was going to have to pay for the airplane time myself. The Wilga's AI-14 engine's starter has a battery operated compressor to keep it charged, so you were still dependent on a battery in the end.  For reliable starting any time anywhere, hard to beat a hand cranked inertia starter.

As far as I know, no modern airliner uses air to power the brakes. The systems are hydraulic.

This makes me wonder if OP heard the term "Air Brake" and assumed it meant pneumatic brakes?


text:
Is the airspeed for the minimum the same for the theoretical power required curve as for the shaft power required curve?
I am aware that the efficiency of the propeller implies a difference in the values of the theoretical power required curve and the corresponding shaft power required, but does that efficiency correction change the abscissae of the respective minima...?
In this diagram (taken from von Mises' 'Fluglehre') the power is specified as 'Zugleistung', i.e. as the theoretical power drag (thrust) x velocity. All usual power required diagrams use that 'theoretical power'. My question is: If we took instead the shaft power, would the airspeed for the minimum of shaft power be the same...?

I think you mean "I am aware that the efficiency of the propeller implies a difference in the values of the theoretical power AVAILABLE curve and the corresponding OUTPUT shaft power"?

No... I mean the power required curve... If you draw the power required curve with the data of the shaft power, you'll get a curve that is different from the one you may construct with a theoretical power that is the product of thrust/drag and airspeed. The reason for the difference is the efficiency of the propeller, that varies with the airspeed. I wonder if the minima of the two curves lie at the same value of airspeed...

What do you mean minima of the two curves? Do you mean the maximum excess power, or the global maximum/minimum of the power available/power required?

In other words: at any point of the power required curve you are in unaccelerated, s/l flight. The power you require for that situation is obviously that of the ordinate of the point of the curve where you are. Now, that power implies a value of shaft power that is different. If, with that data for different airspeeds, you build a 'shaft power required curve', would the minima of both curves be the same...?

It would have to be a large change in efficiency to change the point at which minimum throttle is required.  Under steady state conditions (where time and distance are constant), mimimum drag airspeed would be minimum (Power) throttle, even though this is not the maximum power available (efficiency) airspeed.

The OP is asking about power required, not power available, which, in my mind, would change the answer.
From an aircraft performance perspective, power required is the product of total aircraft drag and inertial speed ($P_R=TV$), applicable only for steady unaccelerated flight. Thus, the minimum of that curve is dependent on profile and vortex drag. In potential flow, it would be where $3C_{D_0}=C_{D_i}$ (see this answer for derivation).
Shaft power required is the power needed to be transferred to the propeller in order to achieve the aforementioned $P_R$. From the actuator disk theory, we know that it must be larger than $P_R$. The shape of this curve would depend also on the propeller itself (e.g. pitch, diameter, airfoil) and its control system (e.g. constant vs variable pitch). Therefore, the minima of the two curves do not have to coincide.
To illustrate this, I've gathered some RC propeller data from APC, which according to its website are generated based on vortex (potential) theory. The drag polar is based on a joke planform designed to accentuate the curvature of the polars:

$C_{D_0}=0.02$ (this is way too low for an RC planform)
$e=0.85$
$W=25lb$
$A=8$
$S_{ref}=15ft^2$

The figure below shows the power required (P_R) for level flight and the shaft power (torque * rotation speed) needed to sustain the level flight for a few prop configurations.

This is good work.  Now, to apply this to full scale, even for an electric motor,  we will have a huge range of RPM to consider as the 16 x 8 is almost 3 times as long and twice as coarse as the 6 x 4.  But what you are showing is these props are more efficient at a speed higher than min sink rate, but the larger, slower one is closer to the power required minimum.  Engine efficiency is part of the explanation for Vy.  Could you run 16 x 4, 16 x 8, 16 x 10?  This would be closer to a variable pitch 25 lb model.  Like the graphs!